Problem 20 - Chemistry: The Central Science

Question

An atom of rhodium ( \(R \mathrm{~h})\) has a diameter of about \(2.7 \times 10^{-8} \mathrm{~cm}\). (a) What is the radius of a rhodium atom in angstroms \((\AA)\) and in meters (m)? (b) How many Rh atoms would have to be placed side by side to span a distance of \(6.0 \mu \mathrm{m}\) ? (c) If you assume that the \(\mathrm{Rh}\) atom is a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single atom?

Step-by-Step Solution

Answer

In summary, the radius of a rhodium atom is approximately \(1.35 \mathrm{~\AA}\) (angstroms) or \(1.35\times 10^{-10} \mathrm{~m}\) (meters). Approximately 22,200 Rh atoms would have to be placed side by side to span a distance of \(6.0 \mu \mathrm{m}\). Assuming the Rh atom is a sphere, the volume of a single atom is approximately \(3.278\times10^{-30} \mathrm{m^3}\).

Step 1: Convert diameter to radius in angstroms and meters

First, let's calculate the radius of the rhodium atom using the formula: radius = diameter × 0.5 Here, the diameter (\(d\)) is \(2.7\times 10^{-8}\mathrm{~cm} \). Now, let's convert the diameter into angstroms and meters: \(1 \mathrm{~cm} = 10^8 \mathrm{~\AA} \) \(1 \mathrm{~cm} = 0.01 \mathrm{~m} \) So, the diameter in angstroms (\(d_\mathrm{\AA}\)) and meters (\(d_\mathrm{m}\)) is: \(d_\mathrm{\AA} = 2.7\times 10^{-8} \mathrm{~cm} \times 10^{8} \mathrm{~\AA/cm} = 2.7 \mathrm{~\AA}\) \(d_\mathrm{m} = 2.7\times 10^{-8} \mathrm{~cm} \times 0.01 \mathrm{~m/cm} = 2.7\times 10^{-10} \mathrm{~m}\) Now, let's find the radius in angstroms (\(r_\mathrm{\AA}\)) and meters (\(r_\mathrm{m}\)): \(r_\mathrm{\AA} = \frac{d_\mathrm{\AA}}{2} = \frac{2.7 \mathrm{~\AA}}{2} = 1.35 \mathrm{~\AA}\) \(r_\mathrm{m} = \frac{d_\mathrm{m}}{2} = \frac{2.7 \times 10^{-10} \mathrm{~m}}{2} = 1.35\times 10^{-10} \mathrm{~m}\)

Step 2: Find the number of Rh atoms required to span 6.0 micrometers

We are given a distance of 6.0 micrometers, and we need to determine how many rhodium atoms are required to cover this distance. To do so, we can divide the distance by the diameter of a single rhodium atom: Number of Rh atoms = Distance / Diameter (in meters) \(1 \mathrm{~\mu m} = 10^{-6} \mathrm{~m}\) Convert the given distance (6.0 μm) to meters: \(6.0 \mathrm{~\mu m} = 6.0\times 10^{-6} \mathrm{~m}\) Now, finding the number of Rh atoms: Number of Rh atoms = \(\frac{6.0\times10^{-6}\mathrm{~m}}{2.7\times10^{-10}\mathrm{~m}} = \frac{6.0}{2.7}\times 10^4 \approx 2.22\times 10^4 \) So, approximately 22,200 rhodium atoms are required to span a distance of 6.0 μm.

Step 3: Compute the volume of a single Rh atom

To compute the volume of a single rhodium atom, we will assume that it is a sphere. The formula for the volume of a sphere is given by: Volume = \(\frac{4}{3}\pi r^3\) We will use the converted radius in meters ($1.35\times 10^{-10}\mathrm{~m}) to calculate the volume of a single Rh atom: Volume = \(\frac{4}{3}\pi (1.35\times10^{-10}\mathrm{~m})^3 = \frac{4}{3}\pi (2.4657\times10^{-30}\mathrm{~m^3}) \approx 3.278\times10^{-30} \mathrm{m^3}\) The volume of a single rhodium atom is approximately \(3.278\times10^{-30} \mathrm{m^3}\).

Key Concepts

Atomic Structure

Atoms are the smallest building blocks of matter, forming the basis of chemistry and physics.
They consist of a nucleus, which contains protons and neutrons, surrounded by a cloud of electrons. The arrangement and number of these particles determine the atom's identity and its chemical properties. The size of an atom is often described using its diameter or radius, which is crucial for calculations in chemistry and physics.
An atom's radius can help infer how atoms bond, how they interact with light, and how they form solids. It's interesting to note that despite the tiny size of atoms, they make up the massive universe around us.

Unit Conversion

Unit conversion is a fundamental skill in science, making it possible to communicate measurements universally. For atomic radius calculations, we need to be familiar with units like centimeters (cm), angstroms (\(\mathrm{\AA}\)), and meters (m).
For instance:

• 1 cm = \(10^8\) Å
• 1 cm = 0.01 m
• 1 \(\mu m\) = \(10^{-6}\) m

These conversions let us express measurements in the most appropriate units for calculations or presentations. Understanding this allows scientists to work with very large or very small numbers efficiently, helping in tasks ranging from scaling distances in microscopy to astronomical calculations.

Volume Calculation

Volume calculation involves determining how much three-dimensional space an object occupies.
When dealing with atoms, which are assumed to be spherical for such calculations, the formula for the volume of a sphere is: \[\text{Volume} = \frac{4}{3}\pi r^3\]Where \(r\) is the radius of the sphere. For the rhodium atom with a radius of \(1.35\times 10^{-10}\) m, you use this formula to find its volume:

• Calculate \(r^3\)
• Multiply by \(\pi\)
• Multiply by \(\frac{4}{3}\)

This gives you the volume of a single rhodium atom. Mastering how to manipulate this formula can help you understand the practical applications of volume in nanotechnology and materials science.

Rhodium Atoms

Understanding rhodium atoms involves recognizing their place on the periodic table and their unique properties. Rhodium (Rh) is a transition metal, known for its high reflectivity and corrosion resistance.
This atomic structure, defined by a specific number of protons and neutrons, gives rhodium its distinctive chemical behavior. It's utilized in industries for catalysts in vehicles and jewelry plating. In scientific calculations, we often assume a spherical shape for rhodium atoms to facilitate mathematical modeling. This simplification lets us make important predictions about how many atoms fit in a specific space or how they contribute to the overall properties of a material. Grasping the basics of rhodium and its atomic arrangement helps in various applications across technology and industry.